A black body radiates energy at the rate of E | vedclass

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(D) According to Stefan-Boltzmann law,the radiant energy $E$ emitted by a black body is directly proportional to the fourth power of its absolute temp

Vedclass · Accepted Answer

$\frac{E}{16}$

Vedclass · Answer

(D) According to Stefan-Boltzmann law,the radiant energy $E$ emitted by a black body is directly proportional to the fourth power of its absolute temperature $T$.Mathematically,$E \propto T^4$.Let $E_1 = E$ at temperature $T_1 = T$.Let $E_2$ be the radiant energy at temperature $T_2 = \frac{T}{2}$.Using the ratio formula: $\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4$.Substituting the values: $\frac{E_2}{E} = \left(\frac{T/2}{T}\right)^4 = \left(\frac{1}{2}\right)^4$.Therefore,$E_2 = \frac{E}{16}$.

$1$. Burning candle	$i$. Line spectrum
$2$. Sodium vapour	$ii$. Continuous spectrum
$3$. Bunsen flame	$iii$. Band spectrum
$4$. Dark lines in solar spectrum	$iv$. Absorption spectrum

$A$ black body radiates energy at the rate of $E \text{ W/m}^2$ at a high temperature $T \text{ K}$. When the temperature is reduced to $\left(\frac{T}{2}\right) \text{ K}$,the radiant energy is

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